Taylor Expansion of sine and cosine

Question

Here: Is there a formula for sine and cosine?, one of the answers mentioned using the Taylor expansion for approximating $\sin(x)$ and $\cos(x)$, and someone commented pointing out however that this would only work for small values of $x$.

My question is, am I mistaken in thinking that the Taylor Series could be used to approximate the sine (and cosine) function anywhere in $\mathbb R $, as long as the formula $\sum_{k=0}^n {f^{(k)}(x_0)\over k!}(x-x_0)^k$ is used, and not the explicit expansion in $0$?

Answer 1

The Taylor expansions of $\sin$ and $\cos$ always converge, also when taking the explicit expansions in the point $0$. What they were referring to on the other page, is: when you only want to approximate, for example, $\sin(x)$, then you could use $$\sin(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}$$ since for small $x$, the terms after that are so small that they are negligible. But for large $x$, this isn't such a good approximation - however, the infinite sum $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$ still converges to the exact value for $\sin(x)$, no matter how big $x$ is.

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Edit. If we take the Taylor expansion in another point than $0$, say, $2\pi$, then we get $$(x-2\pi)-\frac{(x-2\pi)^3}{3!}+\frac{(x-2\pi)^5}{5!}-\cdots$$ As we can see, taking only the first three terms would approximate $\sin(x)$ differently - it working better for values close to $2\pi$. This way we can intuitively see that yes, taking the Taylor expansion on another point does influence the way $\sin(x)$ approximates $x$.